Metal constructions is a complex and extremely responsible topic. Even a small mistake can cost hundreds of thousands and millions of dollars. In some cases, the price of a mistake can be the lives of people at a construction site, as well as during operation. So, checking and rechecking calculations is necessary and important.

Using Excel to solve calculation problems is, on the one hand, not a new thing, but at the same time not quite familiar. However, Excel calculations have a number of undeniable advantages:

• openness- each such calculation can be disassembled by bones.
• Availability- the files themselves exist in the public domain, are written by the developers of the MK to suit their needs.
• Convenience- almost any PC user is able to work with programs from the MS Office package, while specialized design solutions are expensive, and, moreover, require serious effort to master.

They should not be considered a panacea. Such calculations make it possible to solve narrow and relatively simple design problems. But they do not take into account the work of the structure as a whole. In a number of simple cases, they can save a lot of time:

• Calculation of a beam for bending
• Calculation of a beam for bending online
• Check the calculation of the strength and stability of the column.
• Check the selection of the bar section.

Universal calculation file MK (EXCEL)

Table for the selection of sections of metal structures, according to 5 different points of SP 16.13330.2011
Actually, using this program, you can perform the following calculations:

• calculation of a single-span hinged beam.
• calculation of centrally compressed elements (columns).
• calculation of stretched elements.
• calculation of eccentric-compressed or compressed-bent elements.

The version of Excel must be at least 2010. To see the instructions, click on the plus in the upper left corner of the screen.

METALLIC

The program is an EXCEL book with macro support.
And it is intended for the calculation of steel structures according to
SP16 13330.2013 "Steel structures"

Selection and calculation of runs

The selection of a run is a trivial task only at first glance. The step of runs and their size depend on many parameters. And it would be nice to have an appropriate calculation at hand. This is what this must-read article is about:

• calculation of a run without strands
• calculation of a run with one strand
• calculation of a run with two strands
• calculation of the run taking into account the bimoment:

But there is a small fly in the ointment - apparently in the file there are errors in the calculation part.

Calculation of the moments of inertia of a section in excel tables

If you need to quickly calculate the moment of inertia of a composite section, or there is no way to determine the GOST according to which the metal structures are made, then this calculator will come to your aid. A little explanation is at the bottom of the table. In general, the work is simple - we select a suitable section, set the dimensions of these sections, and obtain the main parameters of the section:

• Moments of inertia of the section
• Section modulus
• Radius of gyration of section
• Cross-sectional area
• static moment
• Distances to the center of gravity of the section.

The table contains calculations for the following types of sections:

• pipe
• rectangle
• I-beam
• channel
• rectangular pipe
• triangle

Often, people who make a covered canopy for a car in the yard or for protection from the sun and precipitation do not calculate the section of the racks on which the canopy will rest, but select the section by eye or after consulting with a neighbor.

You can understand them, the loads on the racks, which in this case are columns, are not so hot, the amount of work performed is also not huge, and the appearance of the columns is sometimes much more important than their bearing capacity, so even if the columns are made with a multiple margin of safety - there is no big problem in it. Moreover, you can spend an infinite amount of time searching for simple and intelligible information about the calculation of solid columns without any result - it is almost impossible to understand the examples of calculating columns for industrial buildings with loads applied at several levels without good knowledge of the strength of materials, and ordering the calculation of the column in an engineering organization can reduce all expected savings to zero.

This article was written with the aim of at least slightly changing the existing state of affairs and is an attempt to simply state the main steps in the calculation of a metal column as simply as possible, nothing more. All basic requirements for the calculation of metal columns can be found in SNiP II-23-81 (1990).

General provisions

From a theoretical point of view, the calculation of a centrally compressed element, which is a column, or a rack in a truss, is so simple that it is even inconvenient to talk about it. It is enough to divide the load by the design resistance of the steel from which the column will be made - that's it. In mathematical terms, it looks like this:

F=N/Ry (1.1)

F- required sectional area of ​​the column, cm²

N- concentrated load applied to the center of gravity of the cross section of the column, kg;

Ry- design resistance of metal to tension, compression and bending in terms of yield strength, kg/cm². The value of the design resistance can be determined from the corresponding table.

As you can see, the level of complexity of the task refers to the second, maximum to the third grade of elementary school. However, in practice, everything is far from being as simple as in theory, for a number of reasons:

1. It is only theoretically possible to apply a concentrated load exactly to the center of gravity of the column cross section. In reality, the load will always be distributed and there will also be some eccentricity of the application of the reduced concentrated load. And if there is an eccentricity, then there is a longitudinal bending moment acting in the cross section of the column.

2. The centers of gravity of the cross sections of the column are located on the same straight line - the central axis, also only theoretically. In practice, due to the inhomogeneity of the metal and various defects, the centers of gravity of the cross sections can be shifted relative to the central axis. And this means that the calculation must be carried out according to the section, the center of gravity of which is as far as possible from the central axis, which is why the eccentricity of the force for this section is maximum.

3. The column may not have a straight shape, but be slightly curved as a result of factory or assembly deformation, which means that the cross sections in the middle part of the column will have the largest load application eccentricity.

4. The column can be installed with deviations from the vertical, which means that a vertically acting load can create an additional bending moment, maximum at the bottom of the column, or more precisely, at the point of attachment to the foundation, however, this is relevant only for free-standing columns .

5. Under the action of the loads applied to it, the column can be deformed, which means that the eccentricity of the load application will again appear and, as a result, an additional bending moment.

6. Depending on how exactly the column is fixed, the value of the additional bending moment at the bottom and in the middle of the column depends.

All this leads to the appearance of a buckling, and the influence of this bending must be somehow taken into account in the calculations.

Naturally, it is practically impossible to calculate the above deviations for a structure that is still being designed - the calculation will be very long, complicated, and the result is still doubtful. But it is very possible to introduce into formula (1.1) a certain coefficient that would take into account the above factors. This coefficient is φ - buckling coefficient. The formula that uses this coefficient looks like this:

F = N/φR (1.2)

Meaning φ is always less than one, this means that the section of the column will always be greater than if you simply calculated using the formula (1.1), this is me to the fact that the most interesting will begin now and remember that φ always less than one - does not hurt. For preliminary calculations, you can use the value φ within 0.5-0.8. Meaning φ depends on steel grade and column flexibility λ :

λ = l ef / i (1.3)

l ef- Estimated length of the column. The calculated and actual length of the column are different concepts. The estimated length of the column depends on the method of fixing the ends of the column and is determined using the coefficient μ :

l ef = μ l (1.4)

l - actual length of the column, cm;

μ - coefficient taking into account the method of fixing the ends of the column. The coefficient value can be determined from the following table:

Table 1. Coefficients μ for determining the effective lengths of columns and racks of constant section (according to SNiP II-23-81 (1990))

As you can see, the value of the coefficient μ varies several times depending on the method of fixing the column, and here the main difficulty is which design scheme to choose. If you don't know which fixing scheme meets your conditions, then take the value of the coefficient μ=2. The value of the coefficient μ=2 is taken mainly for free-standing columns, a good example of a free-standing column is a lamppost. The value of the coefficient μ=1-2 can be taken for canopy columns on which beams are supported without rigid attachment to the column. This design scheme can be accepted when the canopy beams are not rigidly attached to the columns and when the beams have a relatively large deflection. If trusses rigidly attached to the column by welding will rest on the column, then the value of the coefficient μ = 0.5-1 can be taken. If there are diagonal ties between the columns, then the value of the coefficient μ = 0.7 can be taken for non-rigid fastening of diagonal ties or 0.5 for rigid fastening. However, such stiffness diaphragms are not always in 2 planes, and therefore such coefficient values ​​should be used with caution. When calculating the racks of trusses, the coefficient μ=0.5-1 is used, depending on the method of fixing the racks.

The value of the coefficient of flexibility approximately shows the ratio of the effective length of the column to the height or width of the cross section. Those. the greater the value λ , the smaller the width or height of the cross section of the column and, accordingly, the greater the margin over the section will be required for the same length of the column, but more on that later.

Now that we have determined the coefficient μ , you can calculate the estimated length of the column using the formula (1.4), and in order to find out the value of the flexibility of the column, you need to know the radius of gyration of the column section i :

where I- the moment of inertia of the cross section relative to one of the axes, and here the most interesting begins, because in the course of solving the problem we just have to determine the required sectional area of ​​the column F, but this is not enough, it turns out, we still need to know the value of the moment of inertia. Since we do not know either one or the other, the solution of the problem is carried out in several stages.

At the preliminary stage, the value is usually taken λ within 90-60, for columns with a relatively small load, λ = 150-120 can be taken (the maximum value for columns is 180, the values ​​of ultimate flexibility for other elements can be found in Table 19 * SNiP II-23-81 (1990). Then according to Table 2, the value of the coefficient of flexibility is determined φ :

Table 2. Buckling coefficients φ of centrally compressed elements.

Note: coefficient values φ in the table are magnified 1000 times.

After that, the required radius of gyration of the cross section is determined by converting formula (1.3):

i = l ef /λ (1.6)

According to the assortment, a rolling profile is selected with the corresponding value of the radius of gyration. Unlike bending elements, where the section is selected only along one axis, since the load acts only in one plane, in centrally compressed columns, longitudinal bending can occur relative to any of the axes, and therefore the closer the value of I z to I y , the better, in other In other words, profiles of round or square section are most preferred. Well, now let's try to determine the section of the column based on the knowledge gained.

An example of the calculation of a metal centrally compressed column

Available: the desire to make a canopy near the house of approximately the following form:

In this case, the only centrally compressed column under any conditions of fastening and under a uniformly distributed load will be the column shown in red in the figure. In addition, the load on this column will be maximum. Columns marked in blue and green in the figure can be considered as centrally compressed, only with an appropriate design solution and a uniformly distributed load, columns marked in orange will be either centrally compressed or eccentrically compressed or frame uprights, calculated separately. In this example, we will calculate the section of the column marked in red. For calculations, we will take a constant load from the own weight of the canopy of 100 kg/m² and a live load of 100 kg/m² from the snow cover.

2.1. Thus, the concentrated load on the column, marked in red, will be:

N = (100+100) 5 3 = 3000 kg

2.2. We take a preliminary value λ = 100, then according to table 2, the bending coefficient φ = 0.599 (for steel with a design strength of 200 MPa, this value is taken to provide an additional margin of safety), then the required sectional area of ​​the column:

F\u003d 3000 / (0.599 2050) \u003d 2.44 cm & sup2

2.3. According to table 1, we accept the value μ \u003d 1 (since the profiled deck roofing, properly fixed, will provide structural rigidity in a plane parallel to the plane of the wall, and in a perpendicular plane, the relative immobility of the top point of the column will ensure the fastening of the rafters to the wall), then the radius of inertia

i= 1 250/100 = 2.5 cm

2.4. According to the assortment for square profile pipes, these requirements are met by a profile with cross-sectional dimensions of 70x70 mm with a wall thickness of 2 mm, having a radius of gyration of 2.76 cm. The cross-sectional area of ​​​​such a profile is 5.34 cm & sup2. This is much more than required by calculation.

2.5.1. We can increase the flexibility of the column, while reducing the required radius of gyration. For example, when λ = 130 bend factor φ = 0.425, then the required sectional area of ​​the column:

F \u003d 3000 / (0.425 2050) \u003d 3.44 cm & sup2

2.5.2. Then

i= 1 250/130 = 1.92 cm

2.5.3. According to the assortment for square profile pipes, these requirements are met by a profile with cross-sectional dimensions of 50x50 mm with a wall thickness of 2 mm, having a radius of gyration of 1.95 cm.

Instead of square profile pipes, you can use an equal-shelf angle, a channel, an I-beam, a regular pipe. If the calculated steel resistance of the selected profile is more than 220 MPa, then the column section can be recalculated. That, in principle, is all that concerns the calculation of metal centrally compressed columns.

Calculation of an eccentrically compressed column

Here, of course, the question arises: how to calculate the remaining columns? The answer to this question depends heavily on how the canopy is attached to the columns. If the canopy beams are rigidly attached to the columns, then a rather complex statically indeterminate frame will be formed, and then the columns should be considered as part of this frame and the section of the columns should be calculated additionally for the action of the transverse bending moment, but we will further consider the situation when the columns shown in the figure , are hinged to the canopy (the column marked in red is no longer considered). For example, the head of the columns has a support platform - a metal plate with holes for bolting the canopy beams. For various reasons, the load on such columns can be transferred with a sufficiently large eccentricity:

The beam shown in the figure, in beige, will bend a little under the influence of the load, and this will lead to the fact that the load on the column will not be transferred along the center of gravity of the column section, but with eccentricity e and when calculating the extreme columns, this eccentricity must be taken into account. There are a great many cases of eccentric loading of columns and possible cross sections of columns, which are described by the corresponding formulas for calculation. In our case, to check the cross section of an eccentrically compressed column, we will use one of the simplest ones:

(N/φF) + (M z /W z) ≤ R y (3.1)

In this case, when we have already determined the section of the most loaded column, it is enough for us to check whether such a section is suitable for the remaining columns, for the reason that we do not have the task of building a steel plant, but we simply calculate the columns for the canopy, which will all be of the same section for reasons of unification.

What's happened N, φ And R we already know.

Formula (3.1) after the simplest transformations will take the following form:

F = (N/R y)(1/φ + e z F/W z) (3.2)

because M z =N e z, why the value of the moment is exactly this and what the moment of resistance W is, is explained in sufficient detail in a separate article.

on the columns indicated in the figure in blue and green, will be 1500 kg. We check the required cross section under such a load and previously determined φ = 0,425

F \u003d (1500/2050) (1 / 0.425 + 2.5 3.74 / 5.66) \u003d 0.7317 (2.353 + 1.652) \u003d 2.93 cm & sup2

In addition, formula (3.2) allows you to determine the maximum eccentricity that the already calculated column can withstand, in this case the maximum eccentricity will be 4.17 cm.

The required cross section of 2.93 cm² is less than the accepted 3.74 cm², and therefore a square profile pipe with a cross section of 50x50 mm and a wall thickness of 2 mm can also be used for the outermost columns.

Calculation of an eccentrically compressed column by conditional flexibility

Oddly enough, but for the selection of the section of an eccentrically compressed column - a solid rod, there is an even simpler formula:

F = N/φ e R (4.1)

φ e- buckling coefficient depending on the eccentricity, it could be called the eccentric buckling coefficient, not to be confused with the buckling coefficient φ . However, the calculation by this formula may be longer than by formula (3.2). To determine the ratio φ e you still need to know the value of the expression e z F/W z- which we met in the formula (3.2). This expression is called relative eccentricity and is denoted m:

m = e z F/W z (4.2)

After that, the reduced relative eccentricity is determined:

m ef = hm (4.3)

h- this is not the height of the section, but a coefficient determined according to table 73 of SNiPa II-23-81. I'll just say that the value of the coefficient h varies from 1 to 1.4, h = 1.1-1.2 can be used for most simple calculations.

After that, you need to determine the conditional flexibility of the column λ¯ :

λ¯ = λ√‾(R y / E) (4.4)

and only after that, according to table 3, determine the value φ e :

Table 3. Coefficients φ e for checking the stability of eccentrically compressed (compressed-bent) solid-walled rods in the plane of action of the moment, coinciding with the plane of symmetry.

Notes:

1. Coefficient values φ are magnified 1000 times.
2. Meaning φ should not be taken more than φ .

Now, for clarity, let's check the section of columns loaded with eccentricity, according to the formula (4.1):

4.1. The concentrated load on the columns marked in blue and green will be:

N \u003d (100 + 100) 5 3/2 \u003d 1500 kg

Load application eccentricity e= 2.5 cm, buckling factor φ = 0,425.

4.2. We have already determined the value of the relative eccentricity:

m = 2.5 3.74 / 5.66 = 1.652

4.3. Now we determine the value of the reduced coefficient m ef :

m ef = 1.652 1.2 = 1.984 ≈ 2

4.4. Conditional flexibility with the coefficient of flexibility adopted by us λ = 130, steel strength R y = 200 MPa and modulus of elasticity E= 200000 MPa will be:

λ¯ = 130√‾(200/200000) = 4.11

4.5. According to table 3, we determine the value of the coefficient φ e ≈ 0.249

4.6. Determine the required section of the column:

F \u003d 1500 / (0.249 2050) \u003d 2.94 cm & sup2

Let me remind you that when determining the cross-sectional area of ​​​​the column using formula (3.1), we got almost the same result.

Advice: In order to transfer the load from the canopy with a minimum eccentricity, a special platform is made in the supporting part of the beam. If the beam is metal, from a rolled profile, then it is usually enough to weld a piece of reinforcement to the bottom flange of the beam.

B-pillar calculation

Racks are called structural elements that work mainly in compression and longitudinal bending.

When calculating the rack, it is necessary to ensure its strength and stability. Ensuring stability is achieved by the correct selection of the section of the rack.

The calculation scheme of the central post is adopted when calculating the vertical load, as hinged at the ends, since it is welded at the bottom and top (see Figure 3).

The B-pillar bears 33% of the total floor weight.

The total weight of the floor N, kg is determined by: including the weight of snow, wind load, load from thermal insulation, load from the weight of the cover frame, load from vacuum.

N \u003d R 2 g,. (3.9)

where g is the total uniformly distributed load, kg / m 2;

R is the inner radius of the tank, m.

The total weight of the floor is made up of the following types of loads:

• 1. Snow load, g 1 . Accepted g 1 \u003d 100 kg / m 2 .;
• 2. Load from thermal insulation, g 2. Accepted g 2 \u003d 45 kg / m 2;
• 3. Wind load, g 3 . Accepted g 3 \u003d 40 kg / m 2;
• 4. Load from the weight of the cover frame, g 4 . Accepted g 4 \u003d 100 kg / m 2
• 5. Taking into account the installed equipment, g 5 . Accepted g 5 \u003d 25 kg / m 2
• 6. Vacuum load, g 6 . Accepted g 6 \u003d 45 kg / m 2.

And the total weight of the overlap N, kg:

The force perceived by the rack is calculated:

The required cross-sectional area of ​​​​the rack is determined by the following formula:

See 2 , (3.12)

where: N is the total weight of the floor, kg;

1600 kgf / cm 2, for steel Vst3sp;

The coefficient of longitudinal bending is structurally accepted = 0.45.

According to GOST 8732-75, a pipe with an outer diameter D h \u003d 21 cm, an inner diameter d b \u003d 18 cm and a wall thickness of 1.5 cm is selected, which is acceptable since the pipe cavity will be filled with concrete.

Pipe cross-sectional area, F:

The moment of inertia of the profile (J), the radius of inertia (r) is determined. Respectively:

J = cm4, (3.14)

where are the geometric characteristics of the section.

Radius of Inertia:

r=, cm, (3.15)

where J is the moment of inertia of the profile;

F is the area of ​​the required section.

Flexibility:

The voltage in the rack is determined by the formula:

kgf/cm (3.17)

At the same time, according to the tables of Appendix 17 (A.N. Serenko) = 0.34

Rack Base Strength Calculation

The design pressure P on the foundation is determined by:

P \u003d P "+ R st + R bs, kg, (3.18)

R st \u003d F L g, kg, (3.19)

R bs \u003d L g b, kg, (3.20)

where: P "-force of the vertical rack P" \u003d 5885.6 kg;

R st - weight racks, kg;

g - specific gravity of steel.g \u003d 7.85 * 10 -3 kg /.

R bs - weight concrete poured into the rack rack, kg;

g b - specific gravity of concrete grade. g b \u003d 2.4 * 10 -3 kg /.

The required area of ​​​​the shoe plate at the allowable pressure on the sandy base [y] f \u003d 2 kg / cm 2:

A slab with sides is accepted: aChb \u003d 0.65×0.65 m. Distributed load, q per 1 cm of the slab is determined:

Estimated bending moment, M:

Estimated moment of resistance, W:

Plate thickness d:

The plate thickness d = 20 mm is taken.

A column is a vertical element of a building's load-bearing structure that transfers loads from higher structures to the foundation.

When calculating steel columns, it is necessary to be guided by SP 16.13330 "Steel structures".

For a steel column, an I-beam, a pipe, a square profile, a composite section of channels, corners, sheets are usually used.

For centrally compressed columns, it is optimal to use a pipe or a square profile - they are economical in terms of metal mass and have a beautiful aesthetic appearance, however, the internal cavities cannot be painted, so this profile must be airtight.

The use of a wide-shelf I-beam for columns is widespread - when the column is pinched in one plane, this type of profile is optimal.

Of great importance is the method of fixing the column in the foundation. The column can be hinged, rigid in one plane and hinged in another, or rigid in 2 planes. The choice of fastening depends on the structure of the building and is more important in the calculation, because. the estimated length of the column depends on the method of fastening.

It is also necessary to take into account the method of attaching purlins, wall panels, beams or trusses to the column, if the load is transferred from the side of the column, then the eccentricity must be taken into account.

When the column is pinched in the foundation and the beam is rigidly attached to the column, the calculated length is 0.5l, but 0.7l is usually considered in the calculation. the beam bends under the action of the load and there is no complete pinching.

In practice, the column is not considered separately, but a frame or a 3-dimensional building model is modeled in the program, it is loaded and the column in the assembly is calculated and the required profile is selected, but in programs it can be difficult to take into account the weakening of the section by bolt holes, so it may be necessary to check the section manually .

To calculate the column, we need to know the maximum compressive / tensile stresses and moments that occur in key sections, for this we build stress diagrams. In this review, we will consider only the strength calculation of the column without plotting.

We calculate the column according to the following parameters:

1. Tensile/compressive strength

2. Stability under central compression (in 2 planes)

3. Strength under the combined action of longitudinal force and bending moments

4. Checking the ultimate flexibility of the rod (in 2 planes)

1. Tensile/compressive strength

According to SP 16.13330 p. 7.1.1 strength calculation of steel elements with standard resistance R yn ≤ 440 N/mm2 in case of central tension or compression by force N should be carried out according to the formula

A n is the cross-sectional area of ​​the net profile, i.e. taking into account the weakening of its holes;

R y is the design resistance of rolled steel (depends on the steel grade, see Table B.5 of SP 16.13330);

γ c is the coefficient of working conditions (see Table 1 of SP 16.13330).

Using this formula, you can calculate the minimum required cross-sectional area of ​​\u200b\u200bthe profile and set the profile. In the future, in the verification calculations, the selection of the section of the column can be done only by the method of selection of the section, so here we can set the starting point, which the section cannot be less than.

2. Stability under central compression

Calculation for stability is carried out in accordance with SP 16.13330 clause 7.1.3 according to the formula

A- the cross-sectional area of ​​​​the gross profile, i.e. without taking into account the weakening of its holes;

R

γ

φ is the coefficient of stability under central compression.

As you can see, this formula is very similar to the previous one, but here the coefficient appears φ , in order to calculate it, we first need to calculate the conditional flexibility of the rod λ (denoted with a dash above).

where R y is the design resistance of steel;

E- elastic modulus;

λ - the flexibility of the rod, calculated by the formula:

where l ef is the calculated length of the rod;

i is the radius of inertia of the section.

Effective lengths l ef columns (pillars) of constant cross section or individual sections of stepped columns in accordance with SP 16.13330 clause 10.3.1 should be determined by the formula

where l is the length of the column;

μ - effective length coefficient.

Effective length factors μ columns (pillars) of constant cross section should be determined depending on the conditions for fixing their ends and the type of load. For some cases of fixing the ends and the type of load, the values μ are shown in the following table:

The radius of gyration of the section can be found in the corresponding GOST for the profile, i.e. the profile must be pre-specified and the calculation is reduced to enumerating the sections.

Because the radius of gyration in 2 planes for most profiles has different values ​​​​on 2 planes (only a pipe and a square profile have the same values) and the fastening can be different, and therefore the calculated lengths can also be different, then the calculation for stability must be made for 2 planes.

So now we have all the data to calculate the conditional flexibility.

If the ultimate flexibility is greater than or equal to 0.4, then the stability coefficient φ calculated by the formula:

coefficient value δ should be calculated using the formula:

odds α And β see table

Coefficient values φ , calculated by this formula, should be taken no more than (7.6 / λ 2) at values ​​of conditional flexibility over 3.8; 4.4 and 5.8 for section types a, b and c, respectively.

For values λ < 0,4 для всех типов сечений допускается принимать φ = 1.

Coefficient values φ are given in Appendix D to SP 16.13330.

Now that all the initial data are known, we calculate according to the formula presented at the beginning:

As mentioned above, it is necessary to make 2 calculations for 2 planes. If the calculation does not satisfy the condition, then we select a new profile with a larger value of the radius of gyration of the section. It is also possible to change the design model, for example, by changing the hinged attachment to a rigid one or by fixing the column in the span with ties, the estimated length of the rod can be reduced.

Compressed elements with solid walls of an open U-shaped section are recommended to be reinforced with planks or gratings. If there are no straps, then the stability should be checked for stability in the bending-torsional form of buckling in accordance with clause 7.1.5 of SP 16.13330.

3. Strength under the combined action of longitudinal force and bending moments

As a rule, the column is loaded not only with an axial compressive load, but also with a bending moment, for example, from the wind. The moment is also formed if the vertical load is applied not in the center of the column, but from the side. In this case, it is necessary to make a verification calculation in accordance with clause 9.1.1 of SP 16.13330 using the formula

where N- longitudinal compressive force;

A n is the net cross-sectional area (taking into account weakening by holes);

R y is the design resistance of steel;

γ c is the coefficient of working conditions (see Table 1 of SP 16.13330);

n, Сx And Сy- coefficients taken according to table E.1 of SP 16.13330

Mx And My- moments about the axes X-X and Y-Y;

W xn,min and W yn,min - section modulus relative to the X-X and Y-Y axes (can be found in GOST on the profile or in the reference book);

B- bimoment, in SNiP II-23-81 * this parameter was not included in the calculations, this parameter was introduced to account for warping;

Wω,min – sectoral section modulus.

If there should be no questions with the first 3 components, then accounting for the bimoment causes some difficulties.

The bimoment characterizes the changes introduced into the linear zones of the stress distribution of the deformation of the section and, in fact, is a pair of moments directed in opposite directions

It is worth noting that many programs cannot calculate the bimoment, including SCAD does not take it into account.

4. Checking the ultimate flexibility of the rod

Flexibility of compressed elements λ = lef / i, as a rule, should not exceed the limit values λ u given in the table

The coefficient α in this formula is the utilization factor of the profile, according to the calculation of the stability under central compression.

As well as the stability calculation, this calculation must be done for 2 planes.

If the profile does not fit, it is necessary to change the section by increasing the radius of gyration of the section or changing the design scheme (change the fastenings or fix with ties to reduce the estimated length).

If the critical factor is the ultimate flexibility, then the steel grade can be taken as the smallest. the steel grade does not affect the ultimate flexibility. The optimal variant can be calculated by the selection method.

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The height of the rack and the length of the arm of the application of the force P is selected constructively, according to the drawing. Let's take the section of the rack as 2Sh. Based on the ratio h 0 /l=10 and h/b=1.5-2, we select a section no more than h=450mm and b=300mm.

Figure 1 - Scheme of loading the rack and cross section.

The total weight of the structure is:

m= 20.1+5+0.43+3+3.2+3 = 34.73 tons

The weight coming to one of the 8 racks is:

P \u003d 34.73 / 8 \u003d 4.34 tons \u003d 43400N - pressure per rack.

The force does not act in the center of the section, so it causes a moment equal to:

Mx \u003d P * L; Mx = 43400 * 5000 = 217000000 (N*mm)

Consider a box-section strut welded from two plates

Definition of eccentricities:

If the eccentricity t x has a value from 0.1 to 5 - eccentrically compressed (stretched) rack; if T from 5 to 20, then the tension or compression of the beam must be taken into account in the calculation.

t x\u003d 2.5 - eccentrically compressed (stretched) rack.

Determining the size of the section of the rack:

The main load for the rack is the longitudinal force. Therefore, to select the section, the calculation for tensile (compressive) strength is used:

From this equation find the required cross-sectional area

,mm 2 (10)

Permissible stress [σ] during endurance work depends on the steel grade, stress concentration in the section, number of loading cycles and cycle asymmetry. In SNiP, the allowable stress during endurance work is determined by the formula

(11)

Design resistance R U depends on the stress concentration and on the yield strength of the material. Stress concentration in welded joints is most often caused by welds. The value of the concentration coefficient depends on the shape, size and location of the seams. The higher the stress concentration, the lower the allowable stress.

The most loaded section of the bar structure designed in the work is located near the place of its attachment to the wall. Attachment with frontal fillet welds corresponds to the 6th group, therefore, RU = 45 MPa.

For the 6th group, with n = 10 -6, α = 1.63;

Coefficient at reflects the dependence of allowable stresses on the cycle asymmetry index p, equal to the ratio of the minimum stress per cycle to the maximum, i.e.

-1≤ρ<1,

as well as from the sign of stresses. Tension promotes, and compression prevents cracking, so the value γ for the same ρ depends on the sign of σ max. In the case of pulsating loading, when σmin= 0, ρ=0 in compression γ=2 in tension γ = 1,67.

As ρ→ ∞ γ→∞. In this case, the allowable stress [σ] becomes very large. This means that the risk of fatigue failure is reduced, but does not mean that strength is ensured, since failure during the first loading is possible. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.

Under static tension (no bending)

[σ] = R y. (12)

The value of the design resistance R y according to the yield strength is determined by the formula

(13)

where γ m is the reliability factor for the material.

For 09G2S σ Т = 325 MPa, γ t = 1,25

In static compression, the allowable stress is reduced due to the risk of buckling:

where 0< φ < 1. Коэффициент φ зависит от гибкости и относительного эксцентриситета. Его точное значение может быть найдено только после определения размеров сечения. Для ориентировочного выбора Атрпо формуле следует задаться значением φ. With a small eccentricity of the load application, φ can be taken = 0.6. This coefficient means that the compressive strength of the rod is reduced to 60% of the tensile strength due to buckling.

We substitute the data in the formula:

Of the two values ​​of [ σ] choose the smallest. And in the future, it will be calculated.

Allowable voltage

Putting the data into the formula:

Since 295.8 mm 2 is an extremely small cross-sectional area, based on the design dimensions and the magnitude of the moment, we increase it to

We will select the channel number according to the area.

The minimum area of ​​​​the channel should be - 60 cm 2

Channel number - 40P. Has options:

h=400 mm; b=115mm; s=8mm; t=13.5mm; F=18.1 cm 2 ;

We get the cross-sectional area of ​​\u200b\u200bthe rack, consisting of 2 channels - 61.5 cm 2.

Substitute the data in formula 12 and calculate the stresses again:

=146.7 MPa

The effective stresses in the section are less than the limiting stresses for the metal. This means that the material of construction can withstand the applied load.

Verification calculation of the overall stability of the racks.

Such a check is required only under the action of compressive longitudinal forces. If forces are applied to the center of the section (Mx=Mu=0), then the reduction in the static strength of the rack due to the loss of stability is estimated by the coefficient φ, which depends on the flexibility of the rack.

The flexibility of the rack relative to the material axis (i.e., the axis intersecting the section elements) is determined by the formula:

(15)

where - the length of the half-wave of the curved axis of the rack,

μ - coefficient depending on the condition of fixing; at console = 2;

i min - radius of inertia, is found by the formula:

(16)

We substitute the data in the formula 20 and 21:

Calculation of stability is carried out according to the formula:

(17)

The coefficient φ y is determined in the same way as with central compression, according to table. 6 depending on the flexibility of the rack λ y (λ yo) when bending around the y axis. Coefficient from takes into account the decrease in stability due to the action of the moment M X.